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Say I have two simple graphs, $A$ and $B$

  • In $A$, I know:

    • one node has 3 nodes at distance of 1, 4 nodes at distance 2, etc.
    • one node has 4 nodes at distance of 1, 1 nodes at distance 2, etc.
    • etc.
  • In $B$, I know:

    • one node has 3 nodes at distance of 1, 4 nodes at distance 2, etc.
    • one node has 4 nodes at distance of 1, 1 nodes at distance 2, etc.
    • etc.

Can I derive that graph $A$ and $B$ are isomorphic to each other? Or is there a counter-example where two graphs look the same from this distance point of view, but are not isomorphic?

It is a necessary condition, so if these simple graphs are isomorphic, they will share these distances. I am wondering if this is a sufficient condition as well.

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  • $\begingroup$ I have indeed corrected the title and added a clarification with what is meant by geodesic distance. $\endgroup$
    – 317070
    Mar 5, 2015 at 16:39
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    $\begingroup$ The complexity of graph isomorphism is a famous open problem in computer science and if your condition were sufficient, that would immediately give a simple polynomial-time algorithm. It's very unlikely that everybody would have missed such a simple algorithm, if one existed. $\endgroup$ Mar 5, 2015 at 20:47
  • $\begingroup$ DRs comment does not formalize this single instance to something more general. a more general way/ algorithm would seem to involve computing a "distance matrix" for the graph. $\endgroup$
    – vzn
    Mar 6, 2015 at 5:41

1 Answer 1

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There are two non-isomorphic graphs with 16 vertices in which each vertex has 6 neighbors and 9 vertices at distance 2: the Shrikhande graph and the $4\times 4$ rook's graph.

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    $\begingroup$ See also strongly regular graphs, which both of these happen to be. $\endgroup$
    – Pål GD
    Mar 6, 2015 at 7:14
  • $\begingroup$ Is the OP's condition sufficient for graphs with less than 16 vertices, then? $\endgroup$
    – BeeOnRope
    Jul 13, 2016 at 23:49
  • $\begingroup$ @BeeOnRope I don't find this a terribly interesting question. $\endgroup$ Jul 13, 2016 at 23:51
  • $\begingroup$ Sorry (I guess?), but It is interesting for me. For example, I want to quickly compute isomorphism groups for graphs for limited n, generally n < 15. So such a shortcut would be useful. In any case, the strength of stackexchange is that it only takes one person to find something interesting to reply! $\endgroup$
    – BeeOnRope
    Jul 13, 2016 at 23:53

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